Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
bool isSwap = false;
int tmp[111][111];
int board[111][111];
queue<pair<int, int> > ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) scanf("%d", &tmp[i][j]);
  if (m >= n) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) board[i][j] = tmp[i][j];
  } else {
    swap(n, m);
    isSwap = true;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) board[i][j] = tmp[j][i];
  }
  for (int i = 0; i < n; i++) {
    int curMin = 999;
    for (int j = 0; j < m; j++) curMin = min(curMin, board[i][j]);
    for (int j = 0; j < curMin; j++) ans.push(make_pair(0, i));
    for (int j = 0; j < m; j++) board[i][j] -= curMin;
  }
  for (int j = 0; j < m; j++) {
    int thisOne = board[0][j];
    for (int i = 0; i < n; i++)
      if (board[i][j] != thisOne) {
        printf("-1\n");
        return 0;
      }
    for (int i = 0; i < thisOne; i++) ans.push(make_pair(1, j));
  }
  printf("%d\n", ans.size());
  while (!ans.empty()) {
    int det = ans.front().first;
    int num = ans.front().second;
    ans.pop();
    if (isSwap) det = 1 - det;
    switch (det) {
      case 0:
        printf("row ");
        break;
      case 1:
        printf("col ");
        break;
    }
    printf("%d\n", num + 1);
  }
  return 0;
}